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11 August, 23:00

A certain medical test is known to detect 72% of the people who are afflicted with the disease Y. If 10 people with the disease are administered the test, what is the probability that the test will show that: All 10 have the disease, rounded to four decimal places? At least 8 have the disease, rounded to four decimal places? At most 4 have the disease, rounded to four decimal places?

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  1. 12 August, 02:41
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    Answer: 10773

    Step-by-step explanation:

    .75^10=0.0563

    look at 8, 9, 10

    8 the probability is 10C8*0.75^8*0.25^2=0.2816

    9 the probability is 0.1877 using the same approach

    At least 8 is 0.5256

    At most 4 is

    0.25^10=essentially 0

    1: 10C1*0.75*0.25^9=<0.0001

    2: 45*0.75^2*0.25^8=0.0004

    3: 120*0.75^3*0.25^7=0.0031

    4: 10C4*0.75^4*0.25^6=0.0162

    That sum is 0.0197
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